Define a stochastic process $(\Phi (t): t\in [0,4])$ by $$\Phi (t) = \left\{ \begin{array}{ll} 2, & \mbox{if $t\in[0,1]$} \\ W(1), & \mbox{if $t\in(1,2]$} \\ W(1.5), & \mbox{if $t\in(2,3]$} \\ 0, & \mbox{if $t\in(3,4]$} \end{array} \right.$$ where W is brownian motion.
I am trying to calculate Variance of the stochastic integral $\int_0^4 \Phi (t) dW(t) $.
Knowing that the expectation of the stochastic integral is 0 $Var(\int_0^4 \Phi (t) dW(t)) = E[(\int_0^4 \Phi (t) dW(t))^2]$
Using Ito Isometry, $E[(\int_0^4 \Phi (t) dW(t))^2] = E[(\int_0^4 \Phi^2 (t) dW(t)]$
So finally, I get $Var(\int_0^4 \Phi (t) dW(t))= E[4W(1) + W^2(1)(W(2)-W(1)) + W^2(1.5)(W(3)-W(2))]$
I haven't been able to proceed from here. Any help would be appreciated.
The Ito isometry states that \begin{align} \mathbb{E}\left[\int_0^t (\phi_s d W_s)^2\right]=\mathbb{E}\left[\int_0^t \phi_s^2 ds\right] \end{align} That is, the integral on the right contains $ds$ rather than $dW_s$. See for example Wikipedia. So then \begin{align} \mathbb{E}\left[\int_0^t \phi_s^2 ds\right] = \mathbb{E}\left[2^2+W_1^2+W_{3/2}^2+0^2\right] \end{align}